In the study of a transistor as an amplifier,if $\alpha = \frac{I_C}{I_E} = 0.98$ and $\beta = \frac{I_C}{I_B} = 49$,where $I_C, I_B,$ and $I_E$ are collector,base,and emitter currents respectively,then $\left(\frac{1}{\alpha} - \frac{1}{\beta}\right)$ is equal to:

If $\alpha = 0.98$ and the current through the emitter $i_e = 20 \, mA$,the value of $\beta$ is

In an $n-p-n$ transistor,$10^{8}$ electrons enter the emitter in $10^{-8} \,s$. If $1\%$ of electrons are lost in the base,the fraction of current that enters the collector and the current amplification factor $\beta$ are respectively:

An $N-P-N$ transistor in a common emitter mode is used as a simple voltage amplifier with a collector current of $4 \ mA$. The terminal of an $8 \ V$ battery is connected to the collector through a load resistance $R_L$ and to the base through a resistance $R_B$. The collector-emitter voltage $V_{CE} = 4 \ V$,base-emitter voltage $V_{BE} = 0.6 \ V$ and base current amplification factor $\beta_{d.c.} = 100$. Calculate the values of $R_L$ and $R_B$.

The emitter-base junction of a transistor is …… biased while the collector-base junction is ……. biased.

In the case of a common emitter transistor amplifier,the ratio of collector current to emitter current is $0.96$. Then,the current gain $(\beta)$ of the amplifier is: